📐 Definite Integrals and Areas

A Level Maths Year 1 · Integration

Level 3 · Ages 16–17

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1 The Fundamental Theorem of Calculus

Imagine the area under a curve $y = \mathrm{f}(x)$, measured from some fixed starting line up to a moving right-hand edge at $x$. Call that area $A(x)$. If the edge moves a tiny distance $\delta x$, the extra strip is almost a rectangle of height $\mathrm{f}(x)$ and width $\delta x$, so

Why area and integration are linked
$\delta A \approx \mathrm{f}(x)\,\delta x \quad\Rightarrow\quad \dfrac{\delta A}{\delta x} \approx \mathrm{f}(x) \quad\Rightarrow\quad \dfrac{dA}{dx} = \mathrm{f}(x)$
The area function $A(x)$ differentiates to give the curve. So $A(x)$ is an integral of $\mathrm{f}(x)$. This is the fundamental theorem of calculus: finding areas and reversing differentiation are the same problem.
The definite integral
$\displaystyle\int_a^b \mathrm{f}(x)\,dx = \Big[\mathrm{F}(x)\Big]_a^b = \mathrm{F}(b) - \mathrm{F}(a)$,  where $\mathrm{F}'(x) = \mathrm{f}(x)$
No $+\,c$. If you included it, you would get $(\mathrm{F}(b) + c) - (\mathrm{F}(a) + c)$, and the $c$ cancels. A definite integral is a number, not a function.
2 Evaluating Definite Integrals
Worked Example 1 — A polynomial

Evaluate $\displaystyle\int_1^3 \left(3x^2 - 2x\right)dx$.

①Integrate: $\left[x^3 - x^2\right]_1^3$
②Upper limit: $27 - 9 = 18$
③Lower limit: $1 - 1 = 0$
④$18 - 0 = 18$
Worked Example 2 — Fractional and negative powers, exact answer

Show that $\displaystyle\int_1^4 \left(\sqrt{x} + \frac{2}{x^2}\right)dx = \frac{37}{6}$.

①Rewrite: $x^{1/2} + 2x^{-2}$, which integrates to $\tfrac23x^{3/2} - 2x^{-1}$.
②$\left[\tfrac23x^{3/2} - \dfrac{2}{x}\right]_1^4$
③Upper: $\tfrac23(8) - \tfrac24 = \tfrac{16}{3} - \tfrac12 = \tfrac{29}{6}$
④Lower: $\tfrac23(1) - 2 = -\tfrac43$
⑤$\tfrac{29}{6} - \left(-\tfrac43\right) = \tfrac{29}{6} + \tfrac{8}{6} = \dfrac{37}{6}$ ✓
"Show that" with an exact answer means a calculator result alone scores nothing. Every line above would be expected in a "show detailed reasoning" question.
Worked Example 3 — An unknown limit

Given that $\displaystyle\int_1^k (2x + 1)\,dx = 10$ and $k > 1$, find $k$.

①$\left[x^2 + x\right]_1^k = \left(k^2 + k\right) - 2$
②$k^2 + k - 2 = 10 \;\Rightarrow\; k^2 + k - 12 = 0$
③$(k + 4)(k - 3) = 0$, so $k = 3$ or $k = -4$.
④Since $k > 1$, $k = 3$.
3 The Area Under a Curve
Area between a curve and the $x$-axis
If $y \geqslant 0$ for $a \leqslant x \leqslant b$, the area is $\displaystyle\int_a^b y\,dx$
Worked Example 4 — Area of a region above the axis

Find the area of the region enclosed by $y = 4x - x^2$ and the $x$-axis.

①Roots: $x(4 - x) = 0$, so $x = 0$ and $x = 4$. The parabola is $\cap$-shaped, so it is above the axis between them.
②$\displaystyle\int_0^4 \left(4x - x^2\right)dx = \left[2x^2 - \tfrac13x^3\right]_0^4$
③$= \left(32 - \tfrac{64}{3}\right) - 0 = \dfrac{32}{3}$
④Area $= \dfrac{32}{3}$ square units.
4 Regions Below the Axis
Where the curve is below the $x$-axis, $y$ is negative, so each thin strip $y\,\delta x$ is negative. The integral comes out negative. Its size is still the area — you just report it as a positive number.
Worked Example 5 — A region below the axis

Find the area of the region between $y = x^2 - 4x + 3$ and the $x$-axis.

①$x^2 - 4x + 3 = (x-1)(x-3)$, so the roots are $1$ and $3$; the curve is below the axis between them.
②$\left[\tfrac13x^3 - 2x^2 + 3x\right]_1^3 = (9 - 18 + 9) - \left(\tfrac13 - 2 + 3\right) = 0 - \tfrac43 = -\tfrac43$
③The area is $\dfrac43$ square units.
Regions on both sides of the axis. If the curve crosses the axis somewhere in $a \leqslant x \leqslant b$, positive and negative parts cancel. Split the integral at each root, integrate each part separately, and add the sizes.
Worked Example 6 — A region that crosses the axis

Find the total area between $y = x^2 - 2x$, the $x$-axis and the lines $x = 0$ and $x = 3$.

①$x^2 - 2x = x(x - 2)$: below the axis for $0 < x < 2$, above it for $2 < x < 3$.
②$\mathrm{F}(x) = \tfrac13x^3 - x^2$, with $\mathrm{F}(0) = 0$, $\mathrm{F}(2) = -\tfrac43$, $\mathrm{F}(3) = 0$.
③$\displaystyle\int_0^2 = -\tfrac43$  (area $\tfrac43$)   and   $\displaystyle\int_2^3 = 0 - \left(-\tfrac43\right) = \tfrac43$
④Total area $= \tfrac43 + \tfrac43 = \dfrac83$ square units.
A single integral $\displaystyle\int_0^3 \left(x^2 - 2x\right)dx = 0$ — the two parts cancel exactly. Zero is obviously not the area.
5 Area Between a Curve and a Line
Area between two graphs
$\displaystyle\text{Area} = \int_a^b \big(\text{upper} - \text{lower}\big)\,dx$,  where $a$ and $b$ are the $x$-coordinates of the intersections
This works wherever the two graphs are — even below the axis — because the height of each thin strip is always top minus bottom. Alternatively, find the area under the upper graph and subtract the area under the lower one (often a trapezium or triangle).
xy 1 2 3 4 5 6 4 8 (4, 8) y = 2x y = 6x − x²
Worked Example 7 — Curve above a line

Find the area of the region enclosed by $y = 6x - x^2$ and $y = 2x$ (shaded above).

①Intersections: $6x - x^2 = 2x \Rightarrow x^2 - 4x = 0 \Rightarrow x = 0$ or $x = 4$.
②Between them the curve is on top (test $x = 1$: $5 > 2$).
③$\displaystyle\int_0^4 \left[(6x - x^2) - 2x\right]dx = \int_0^4 \left(4x - x^2\right)dx$
④$= \left[2x^2 - \tfrac13x^3\right]_0^4 = 32 - \tfrac{64}{3} = \dfrac{32}{3}$
Alternative: area under the curve from $0$ to $4$ is $\left[3x^2 - \tfrac13x^3\right]_0^4 = 48 - \tfrac{64}{3} = \tfrac{80}{3}$; the triangle under the line has area $\tfrac12 \times 4 \times 8 = 16$; and $\tfrac{80}{3} - 16 = \tfrac{32}{3}$ ✓
Worked Example 8 — A line above a curve

Find the area of the region enclosed by $y = x^2 + 1$ and $y = 7 - x$.

①$x^2 + 1 = 7 - x \Rightarrow x^2 + x - 6 = 0 \Rightarrow (x+3)(x-2) = 0$, so $x = -3$ or $x = 2$.
②At $x = 0$ the line gives $7$ and the curve $1$, so the line is on top.
③$\displaystyle\int_{-3}^{2}\left(6 - x - x^2\right)dx = \left[6x - \tfrac12x^2 - \tfrac13x^3\right]_{-3}^{2}$
④Upper: $12 - 2 - \tfrac83 = \tfrac{22}{3}$.   Lower: $-18 - \tfrac92 + 9 = -\tfrac{27}{2}$.
⑤$\tfrac{22}{3} + \tfrac{27}{2} = \tfrac{44}{6} + \tfrac{81}{6} = \dfrac{125}{6}$ square units.
With a negative lower limit, put the substitution in brackets: $(-3)^3 = -27$, so $-\tfrac13(-3)^3 = +9$.
6 Quick Reference

Fundamental theorem

$\dfrac{dA}{dx} = \mathrm{f}(x)$: area is found by integrating.

Definite integral

$\int_a^b \mathrm{f}(x)\,dx = \mathrm{F}(b) - \mathrm{F}(a)$.

No constant

$c$ cancels in a definite integral.

Above the axis

Area $= \int_a^b y\,dx$.

Below the axis

Integral is negative; area is its size.

Crossing the axis

Split at the roots; add the sizes.

Between graphs

$\int_a^b (\text{upper} - \text{lower})\,dx$.

Limits

Intersection $x$-values; solve the equations simultaneously.

Negative limits

Use brackets when substituting.

7 Practice Questions
Question 1

Evaluate $\displaystyle\int_0^2 \left(3x^2 + 1\right)dx$.

▶ Show solution

$\left[x^3 + x\right]_0^2 = (8 + 2) - 0 = 10$

Question 2

Evaluate $\displaystyle\int_1^2 \left(4x^3 - 6x\right)dx$.

▶ Show solution

$\left[x^4 - 3x^2\right]_1^2 = (16 - 12) - (1 - 3) = 4 + 2 = 6$

Question 3

Evaluate $\displaystyle\int_1^9 \frac{1}{\sqrt{x}}\,dx$.

▶ Show solution

$x^{-1/2}$ integrates to $2x^{1/2}$.

$\left[2\sqrt{x}\right]_1^9 = 6 - 2 = 4$

Question 4

Find the area between the curve $y = x^2 + 2$, the $x$-axis and the lines $x = 1$ and $x = 3$.

▶ Show solution

$x^2 + 2 > 0$ everywhere, so the integral gives the area directly.

$\left[\tfrac13x^3 + 2x\right]_1^3 = (9 + 6) - \left(\tfrac13 + 2\right) = 15 - \tfrac73 = \dfrac{38}{3}$

Question 5

Find the area of the region enclosed by $y = 9 - x^2$ and the $x$-axis.

▶ Show solution

Roots at $x = \pm 3$; the curve is above the axis in between.

$\left[9x - \tfrac13x^3\right]_{-3}^{3} = (27 - 9) - (-27 + 9) = 18 + 18 = 36$

Question 6

Find the area of the region between $y = x^2 - 6x + 5$ and the $x$-axis.

▶ Show solution

$x^2 - 6x + 5 = (x-1)(x-5)$: roots $1$ and $5$, curve below the axis between them.

$\left[\tfrac13x^3 - 3x^2 + 5x\right]_1^5 = \left(\tfrac{125}{3} - 75 + 25\right) - \left(\tfrac13 - 3 + 5\right) = -\tfrac{25}{3} - \tfrac73 = -\tfrac{32}{3}$

Area $= \dfrac{32}{3}$ square units.

Question 7

Evaluate $\displaystyle\int_1^4 \frac{x^2 + 2}{x^2}\,dx$.

▶ Show solution

Split: $1 + 2x^{-2}$, which integrates to $x - 2x^{-1}$.

$\left[x - \dfrac2x\right]_1^4 = \left(4 - \tfrac12\right) - (1 - 2) = \tfrac72 + 1 = \dfrac92$

Question 8

Given that $\displaystyle\int_0^a \left(6x^2 - 4x\right)dx = 36$, show that $a^3 - a^2 - 18 = 0$, and hence find $a$.

▶ Show solution

$\left[2x^3 - 2x^2\right]_0^a = 2a^3 - 2a^2 = 36$, so $a^3 - a^2 - 18 = 0$ ✓

Try $a = 3$: $27 - 9 - 18 = 0$, so $(a - 3)$ is a factor.

$a^3 - a^2 - 18 = (a - 3)\left(a^2 + 2a + 6\right)$, and $a^2 + 2a + 6$ has discriminant $4 - 24 < 0$, so no other real roots.

$a = 3$

Question 9

Find the area of the region enclosed by the curve $y = x^2$ and the line $y = 2x + 3$.

▶ Show solution

$x^2 = 2x + 3 \Rightarrow x^2 - 2x - 3 = 0 \Rightarrow (x-3)(x+1) = 0$, so $x = -1$ or $x = 3$.

At $x = 0$ the line ($3$) is above the curve ($0$).

$\displaystyle\int_{-1}^{3}\left(2x + 3 - x^2\right)dx = \left[x^2 + 3x - \tfrac13x^3\right]_{-1}^{3}$

$= (9 + 9 - 9) - \left(1 - 3 + \tfrac13\right) = 9 + \tfrac53 = \dfrac{32}{3}$

Question 10

The curve $C$ has equation $y = x^3 - 4x^2 + 3x$.

(a) Find the points where $C$ meets the $x$-axis.   (b) Evaluate $\displaystyle\int_0^3 \left(x^3 - 4x^2 + 3x\right)dx$.   (c) Find the total area of the regions enclosed by $C$ and the $x$-axis.   (d) Explain why your answers to (b) and (c) are different.

▶ Show solution

(a) $x^3 - 4x^2 + 3x = x(x^2 - 4x + 3) = x(x - 1)(x - 3)$, so $C$ meets the axis at $(0, 0)$, $(1, 0)$ and $(3, 0)$.

(b) $\mathrm{F}(x) = \tfrac14x^4 - \tfrac43x^3 + \tfrac32x^2$.

$\mathrm{F}(3) = \tfrac{81}{4} - 36 + \tfrac{27}{2} = -\tfrac94$ and $\mathrm{F}(0) = 0$, so the integral is $-\tfrac94$.

(c) A positive cubic with roots $0, 1, 3$ is above the axis for $0 < x < 1$ and below it for $1 < x < 3$.

$\mathrm{F}(1) = \tfrac14 - \tfrac43 + \tfrac32 = \tfrac{3 - 16 + 18}{12} = \tfrac{5}{12}$, so $\displaystyle\int_0^1 = \tfrac{5}{12}$.

$\displaystyle\int_1^3 = -\tfrac94 - \tfrac{5}{12} = -\tfrac{27}{12} - \tfrac{5}{12} = -\tfrac{8}{3}$, area $\tfrac83$.

Total area $= \tfrac{5}{12} + \tfrac{32}{12} = \dfrac{37}{12}$ square units.

(d) In (b) the region below the axis contributes a negative amount, which partly cancels the positive part: $\tfrac{5}{12} - \tfrac{8}{3} = -\tfrac94$. Area is always positive, so the parts must be found separately and their sizes added.

Definite Integrals and Areas · A Level Maths Year 1 · Created with MathJax